As a part of a customer loyalty program, a restaurant's computer chooses orders at random to receive a free appetizer. Each order has a $1$ in $18$ chance of receiving a free appetizer. (Each order is limited to one free appetizer.) Let $X$ be the number of orders the restaurant fills in a day until they give away the first appetizer. Assume that each order getting the appetizer is independent. Find the probability that the restaurant first gives away an appetizer on the $6^{\text{th}}$ order of the day. You may round your answer to the nearest hundredth. $P(X=6)=$
Solution: Without a fancy calculator For each order: $P({\text{free appetizer}})=\dfrac{1}{18}$ $P(\text{not}})=\dfrac{17}{18}$ If the restaurant first gives away an appetizer on the $6^{\text{th}}$ order, there must be $5$ orders without a free appetizer followed by $1$ with the free appetizer. $\begin{aligned} P(X=6)&=P(\text{NNNNN}}{\text{F}}) \\\\ &=\left(\dfrac{17}{18}}\right)\left(\dfrac{17}{18}}\right)\left(\dfrac{17}{18}}\right)\left(\dfrac{17}{18}}\right)\left(\dfrac{17}{18}}\right)\left({\dfrac{1}{18}}\right) \\\\ &=\left(\dfrac{17}{18}\right)^5\left(\dfrac{1}{18}\right) \\\\ &\approx 0.0417 \end{aligned}$ $P(X=6) \approx 0.0417$